Subnet division. An institution is given the following IP address block by its ISP: 18.14.8.0/21. Divide this IP address block into smaller subnets to fit for three departments, which require 1000, 200, and 200 IP addresses. To receive full credit, you must: 1. Start from the 1000-host subnet. 2. Make sure the three subnets are continuous. 3. Show your work, and fill in the final answer. Also, consider x.x.x.0 and x.x.x.255 as valid IPs. (A) Show your work (B) For the 1000-host department: 1. Subnet address (x.x.x.x/x):. 2. Subnet mask: 3. First IP: 4. Last IP: (C) For the first 200-host department: 1. Subnet address (x.x.x.x/x):. 2. Subnet mask: 3. First IP: 4. Last IP: (D) For the second 200-host department:

Subnet division. An institution is given the following IP address block by its ISP: 18.14.8.0/21. Divide this IP address block into smaller subnets to fit for three departments, which require 1000, 200, and 200 IP addresses. To receive full credit, you must: 1. Start from the 1000-host subnet. 2. Make sure the three subnets are continuous. 3. Show your work, and fill in the final answer. Also, consider x.x.x.0 and x.x.x.255 as valid IPs. (A) Show your work (B) For the 1000-host department: 1. Subnet address (x.x.x.x/x):. 2. Subnet mask: 3. First IP: 4. Last IP: (C) For the first 200-host department: 1. Subnet address (x.x.x.x/x):. 2. Subnet mask: 3. First IP: 4. Last IP: (D) For the second 200-host department:

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

dont dont dont answer existing one and answer only you are 100% sure else directly report the answer
Question 4: The block of IP address 213.40.0.0/16 is granted to an organisation. The administrator is planning to create 1024 subnets. In subnet 1023, the first and last addresses are?
dont post existing one dont dont dont answer if you are not sure
Subject - Data communication networks How can you write the following IPv4 address 192.168.10.2 in IPv6 format? A network on the Internet has a subnet mask of 255.255.248.0. What is the maximum number of hosts it can handle? Explain two problems of SMTP. Thank you in advance!
Question 1: Create subnetworks for the following given requirements (IPV4 addresses). 191.148.1.0/24 – Create VLSM scheme for two subnets, Subnet A requires 30 hosts and Subnet B needs 10. 157.149.30.0/24 – Create VLSM scheme for two subnets, Subnet A requires 5 hosts and Subnet B needs 2 hosts Question 2: Research Security issues in IoT (Internet of Things) (one-page font size should be 12 and font style Times New Roman, line spacing single)
Please answer all the questions. To answer the following question please use the following IP address scheme. The Given IP address is: 129.238.16.154 and subnet Mask 255.255.255.248. 1. What is number of Hosts a) 8 b) 6 c) 16 d) 14 2. What is the first usable IP Address a)b129.238.16.1 b) 129.238.16.152 c) 129.238.16.155 d) 129.238.16.153 3. The CIDAR IP addressing scheme representation is a) 129.238.16.154/248 b) 129.238.16.154/29 c)129.238.16.154/23 d)129.238.16.154/24 4) What is the class of the given IP Address a) Class A b) Class C c) Class B d) Class D 5) The broadcast IP address is a) 129.238.16.255 b)129.238.16.159 c) 129.238.16.248 d)129.238.16.254 6) The number of Subnets is a) 30 b) 32 c) 14 d) 64
I need the answer 20 min
DO NOT COPY FROM OTHER WEBSITES
Please answer 4-6 ASAPPPPPP Please make sure to be organized and say which is which.
IP addressing will have significant impacts on the future extension of the network. You need to decide whether you will be using IPv4 or IPv6. Whatever you are choosing, you need to justify your selection. You need to subnet the networks in an efficient way to ensure optimal use of available IP addresses. In this section, you need to provide a table of IP address and subnet masks for each device of the network.